JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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Analyzing +4 | -1

The minimum value of

{x^2} + \frac{1}{{1 + {x^2}}}\;

is at

View answer and explanation

Correct option: A

x = 0

Option A (

x = 0

) is correct because the function's derivative is zero at

x=0

, indicating a minimum value.

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f(x)=\left(1+b^2\right) x^2+2 b x+1

m(b)

f(x)

b

m(b)

f: R \rightarrow R

f(x)=(x-3)^{n_1}(x-5)^{n_2}, n_1, n_2 \in N

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x<0

x>0

p

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\left[ {1/\sqrt 3 ,\;\sqrt 3 } \right]

f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right),

\frac{k}{e}

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